Decidability and Complexity of Model Checking

We can immediately conclude that model-checking graded strategy logic is decidable. Theorem 4.3.3 The model-checking problem for GRADEDSL is decidable. Moreover, the complexity is not bounded by any fixed tower of exponentials.

Proof.For decidability, use Lemma 4.3.1 to transform the arena and ϕ into an APTand test its emptiness. The lower bound already holds for SL [MMPV14].  In the rest of this section we supply a finer analysis of the complexity of various fragments of graded strategy logic. To do this, we first analyse the number of states of the APT

4.3. Model-checking GRADEDSL

constructed in Lemma 4.3.1.

All the cases in the induction incur at most a linear blowup except for the quantification case. For the quantification case, in case g ∈ {ℵ0, ℵ1, 2ℵ0} the blowup is non-elementary.

In case g ∈ N then the translation incurs an exponential blowup. Indeed, the number of states of the APTBM,X is g × n times the number of states of the APTfor ψ, and since CX

consists of the conjunction of g(g − 1) automata (one for each pair of tuples), and each such automaton has O(n) many states, the number of states of CX is O(ng2). Thus, the number of

states of the APTDM,X is polynomial in the number of states of the APTfor ψ. Finally, the

translation from an APTto an NPTresults in an exponentially larger automaton [KVW00].

In case all grades are from N and the formulas are written using the universal-strategy quantifier shorthand, we can easily modify the construction to handle quantifier-blocks in one shot as if they were a single quantifier, i.e., with a single exponential blowup. For instance, suppose φ = hhy1, . . . , ymii≥hhhx1, . . . , xnii≥gψ (additional quantifiers are treated

similarly). As in the proof, let M be the APT for ψ and take DM,X. Now, instead of

immediately removing alternation and projecting, build DM0_,Y where M0 is D_M,X and

Y , {yji : i ≤ m, j ≤ h}. Finally, remove alternation from DM0,Y to get an NPT

N0_{, and then apply the (X ∪ Y )-projection to the language of N}0 _{to get the desired A}

PT

for φ. Note that the size of DM0_,Y is exponential in the number of states of M since

the costly step of removing alternation is performed only once. Similarly, to deal with a block of universal strategy quantifiers simply use dualisation. For instance, to deal with φ = [[y1, . . . , ym]]<h[[x1, . . . , xn]]<gψ apply the previous procedure to the equivalent formula

¬hhy₁, . . . , ymii≥hhhx1, . . . , xnii≥g¬ψ (recall that negating an APTis done by dualisation,

which incurs no blowup).

Recall that GRADED_NSL denotes the fragment of GRADEDSL in which all grades are restricted to N.

Theorem 4.3.4

1. For everyk ≥ 1, the model-checking problem for GRADED_NSL formulas of quantifier- block rank at mostk is in (k + 1)EXPTIME.

2. For everyk ≥ 1, the model-checking problem for GRADED_NSL formulas of the form ϕ = ℘ψ, where ℘ is a quantifier-block of ϕ and ψ is of quantifier-block rank at most k − 1, is in kEXPTIME.

3. The model-checking problem forGRADED_NSL formulas of the form hhx1, . . . , xnii≥gψ

or[[x1, . . . , xn]]<gψ is in EXPTIMEw.r.t. the parameterg (written in unary).

Proof.Before proving the upper bounds, recall that the complexity of checking emptiness (resp. universality) of an APTis in EXPTIMEin the number of states [KVW00].

For item 1 proceed as follows. As discussed after Lemma 4.3.1, for the case that all grades are in N, the number of states of the APTfor ϕ is a tower of exponentials whose height is the quantifier-block rank of ϕ. This gives the (k + 1)EXPTIMEbound.

For item 2 suppose that ϕ = ℘ψ where ℘ consists of, say, n existential quantifiers (resp. universal quantifiers). The quantifier-block rank of ψ is k − 1. Moreover, the APTDψ, whose

number of states is non-elementary in k − 1, has the property that it is non-empty (resp. universal) if and only if the arena satisfies ℘ψ. Conclude that model checking ℘ψ can be solved in kEXPTIME.

For item 3 first observe that the size of the APT constructed in Lemma 4.3.1 grows quadratically in g. The statement follows by recalling that the complexity of model-checking formulas of this form is exponential in the number of states of the APT. 

Theorem 4.3.5 For every k ≥ 1, the model-checking problem for GRADED_NSL[NG]formulas of alternation number at mostk is in (k + 1)EXPTIMEandkEXPSPACE-hard.

Proof.The lower bound already holds for SL[NG][MMPV14]. For the upper bound, as for SL[NG], note that principal formulas are “state formulas”, i.e., their truth value only depends on the state in which they are interpreted (this is because they have no free placeholders). Thus, one can apply the following marking algorithm to a formula Φ. For every principal subformula ϕ = ℘ψ of Φ for which ψ is quantifier-free, and for every state s of A, introduce a new atomic proposition pϕ. Label s by pϕ(i.e., extend the labelling function at s to include

pϕ) iff A, s |= ϕ. Replace the subformula ϕ of Φ by the atom pϕ, and repeat this process.

Observe that the complexity of marking a state is in (k + 1)EXPTIME where k = alt(ϕ) (indeed, alt(ϕ) = qbr(ϕ) since ϕ = ℘ψ and ψ is quantifier-free, and thus one can apply Theorem 4.3.4 item 1). Also, the cost of the whole marking algorithm is the sum of the costs of all the marking rounds, and the number of rounds is at most the size of the formula. After the marking algorithm has completed, one is left with a Boolean combination of atomic propositions which is trivial to evaluate at each state. Thus the total time is at most

(k + 1)EXPTIME. 

Theorem 4.3.6 The model-checking problem for GRADED_NSL[1G] is 2EXPTIME- COMPLETE.

Proof. The lower-bound already holds for SL[1G][MMPV14]. The upper-bound follows the same algorithm as for SL[1G][MMPV14] but uses the (no more complex) construction from

Lemma 4.3.1 for the strategic quantifier hhx1, . . . , xnii≥ginstead of hhxii. 

Finally, since model checking ATL? is 2EXPTIME-HARD, and ATL? is a syntactic fragment of GRADED_NATL?, which itself is a syntactic fragment of GRADED_NSL[1G], we